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The range of integers that can be represented by an $n$ bit $2’s$ complement number system is:

  1. $-2^{n-1} \text{ to } (2^{n-1} -1)$

  2. $-(2^{n-1} -1) \text{ to } (2^{n-1} -1)$

  3. $-2^{n-1} \text{ to } 2^{n-1}$

  4. $-(2^{n-1} +1) \text{ to } (2^{n-1} -1)$

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Answer: A
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Take example

-128 to +127

0 is occuring 0 time

hence this is the exact way to represent range of numbers in 2’s complement form

 

Hence Option A is the correct answer.
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Best answer
Total number of distinct numbers that can be represented using $n$ bits $=2^n.$

In case of unsigned numbers these corresponds to numbers from $0$ to $2^n -1.$

In case of signed numbers in $1's$ complement or sign magnitude representation, these corresponds to numbers from $-(2^{n-1}-1)$ to $2^{n-1}-1$ with $2$ separate representations for $0.$

In case of signed numbers in $2's$ complement representation, these corresponds to numbers from $-2^{n-1}$ to $2^{n-1}-1$ with a single representation for $0.$
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An n-bit two's-complement numeral system can represent every integer in the range −(2n − 1) to +(2n − 1 − 1).

while ones' complement can only represent integers in the range −(2n − 1 − 1) to +(2n − 1 − 1).

A is answer

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option a

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